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Date May Example questions Marks available 4 Reference code EXM.3.AHL.TZ0.4
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Find Question number 4 Adapted from N/A

Question

This question investigates some applications of differential equations to modeling population growth.

One model for population growth is to assume that the rate of change of the population is proportional to the population, i.e. dPdt=kP, where kR, t is the time (in years) and P is the population

The initial population is 1000.

Given that k=0.003, use your answer from part (a) to find

Consider now the situation when k is not a constant, but a function of time.

Given that k=0.003+0.002t, find

Another model for population growth assumes

Show that the general solution of this differential equation is P=Aekt, where AR.

[5]
a.

the population after 10 years

[2]
b.i.

the number of years it will take for the population to triple.

[2]
b.ii.

limtP

[1]
b.iii.

the solution of the differential equation, giving your answer in the form P=f(t).

[5]
c.i.

the number of years it will take for the population to triple.

[4]
c.ii.

Show that dPdt=mLP(LP), where mR.

[2]
d.

Solve the differential equation dPdt=mLP(LP), giving your answer in the form P=g(t).

[10]
e.

Given that the initial population is 1000, L=10000  and m=0.003, find the number of years it will take for the population to triple.

[4]
f.

Markscheme

1PdP=kdt        M1A1

lnP=kt+c         A1A1

P=ekt+c         A1

P=Aekt, where A=ec         AG

[5 marks]

a.

when t=0,P=1000

A=1000         A1

P(10)=1000e0.003(10)=1030         A1

[2 marks]

b.i.

3000=1000e0.003t        M1

t=ln30.003=366 years        A1

[2 marks]

b.ii.

limtP=       A1

[1 mark]

b.iii.

1PdP=(0.003+0.002t)dt        M1

lnP=0.003t+0.001t2+c        A1A1

P=e0.003t+0.001t2+c        A1

when t=0,P=1000

ec=1000        M1

P=1000e0.003t+0.001t2

[5 marks]

c.i.

3000=1000e0.003t+0.001t2        M1

ln3=0.003t+0.001t2        A1

Use of quadratic formula or GDC graph or GDC polysmlt        M1

t=31.7 years         A1

[4 marks]

c.ii.

k=m(1PL) , where m is the constant of proportionality        A1

So dPdt=m(1PL)P        A1

dPdt=mLP(LP)        AG

[2 marks]

d.

1P(LP)dP=mLdt        M1

1P(LP)=AP+BLP        M1

1A(LP)+BP        A1

A=1L,B=1L        A1

1L(1P+1LP)dP=mLdt

1L(lnPln(LP))=mLt+c        A1A1

ln(PLP)=mt+d, where d=cL        M1

PLP=Cemt, where C=ed        A1

P(1+Cemt)=CLemt        M1

P=CLemt(1+Cemt) (=L(Demt+1),whereD=1C)        A1

[10 marks]

e.

1000=10000D+1        M1

D=9        A1

3000=100009e0.003t+1        M1

t=450 years        A1

[4 marks]

f.

Examiners report

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a.
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b.i.
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b.ii.
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b.iii.
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c.i.
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c.ii.
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d.
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e.
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f.

Syllabus sections

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